Exercise 19 Page 531

The shortest distance between a point and a line is the length of the line segment that is perpendicular to the given line. We will need to find the perpendicular line and then we can find the intersection point. Finally, we can calculate the distance from the given point to the point of intersection.

Finding the Perpendicular Line

Perpendicular lines have negative reciprocal slopes. This means the product of their slopes is equal to -1. m_1 * m_2=- 1 Examining the slope-intercept form of the line representing the nature trail, we know that it has a slope of 13. By substituting this value into the formula, we can find the slope of the perpendicular line. The slope of the perpendicular line is - 3. To find the equation of the line, we substitute the known point in the slope-intercept form, setting the slope to - 3.

y=- 3x+b

SubstituteII

x= - 6, y= 4

4=- 3( - 6)+b

▼

Solve for b

MultNegNegOnePar

- a(- b)=a* b

4=18+b

SubEqn

LHS-18=RHS-18

- 14=b

RearrangeEqn

Rearrange equation

b=- 14

The equation of the perpendicular line is y=- 3x-14

Finding the Point of Intersection

To find our distance, we also need to know where the given line and the perpendicular line intersect. By setting up a system of equations, we can find the point of intersection. y=- 3x-14 y= 13x-4 Since both equations have y isolated, it's convenient to use the Substitution Method.

y=- 3x-14 & (I) y= 13x-4 & (II)

Substitute

(II): y= - 3x-14

y=- 3x-14 - 3x-14= 13x-4

▼

(II): Solve for x

AddEqn

(II): LHS+4=RHS+4

y=- 3x-14 - 3x-10= 13x

MultEqn

(II): LHS * 3=RHS* 3

y=- 3x-14 3(- 3x-10)=x

Distr

(II): Distribute 3

y=- 3x-14 - 9x-30=x

AddEqn

(II): LHS+9x=RHS+9x

y=- 3x-14 - 30=10x

DivEqn

(II): .LHS /10.=.RHS /10.

y=- 3x-14 - 3=x

RearrangeEqn

(II): Rearrange equation

y=- 3x-14 x=- 3

Having solved the second equation for x, we can substitute this value into the first equation to find the value of y.

y=- 3x-14 x=- 3

Substitute

(I): x= - 3

y=- 3( - 3)-14 x=- 3

MultNegNegOnePar

(I): - a(- b)=a* b

y=9-14 x=- 3

SubTerm

(I): Subtract term

y=- 5 x=- 3

The lines intersect at (- 3,- 5).

Finding the Distance

When we know the coordinates of the two points making up the segment, we can use the Distance Formula to calculate it.

d = sqrt((x_2-x_1)^2 + (y_2-y_1)^2)

SubstitutePoints

Substitute ( - 3,- 5) & ( - 6,4)

d=sqrt(( - 3-( - 6))^2+( - 5- 4)^2)

▼

Simplify left-hand side

SubNeg

a-(- b)=a+b

d=sqrt((- 3+6)^2+(- 5-4)^2)

AddSubTerms

Add and subtract terms

d=sqrt(3^2+(- 9)^2)

CalcPow

Calculate power

d=sqrt(9+81)

AddTerms

Add terms

d=sqrt(90)

The distance from the point (- 6,4) to the line y=x+4 is sqrt(90).

Since each unit in the coordinate plane represents 10 feet, we find the total distance by multiplying the unit distance by the distance between the points. 10* sqrt(90)≈ 94.9 feet